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# The four distinct points (0,0), (2,0), (0,-2) and (k, -2) are concyclic, if k is equal to

Vedant
35 Points
3 years ago
Let point (0,0) be a, point(2,0) be b and point(0,-2) be c

A given number of points are said to be concyclic if they lie on the same circle
i.e. they all are equidistant from a given point

We know that the vertices of a square are equidistant from the centre of the square
Therefore, they are concyclic

On plotting the points a, b, c , we come to know points a and b are equidistant from each other
Also points a and c are equidistant from each other and angle(bac) is a right angle

We can see that our points combine to form an isosceles right triangle (which is half of a square)
To make points a, b, c and (k, -2) concyclic we assume them to be the vertices of a square

Let us see if our condition is possible

The difference between the y values of c and a is 2
The difference between the x values of a and b is 2
The difference between the y values of b and (k, -2) is 2
$\therefore$ The difference between the x values of (k, -2) and c MUST BE 2
$\Rightarrow$ |k – 0| = 2
k – 0 = 2
k = 2
Regards
Vedant